/*
给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words，找出所有同时在二维网格和字典中出现的单词。

单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

 

示例 1：


输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]
示例 2：


输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]
 

提示：

m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] 是一个小写英文字母
1 <= words.length <= 3 * 104
1 <= words[i].length <= 10
words[i] 由小写英文字母组成
words 中的所有字符串互不相同

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/

#include "../stdc++.h"

// 回溯 + 字典树
struct TrieNode {
    string word;
    unordered_map<char,TrieNode *> children;
    TrieNode() {
        this->word = "";
    }   
};

void insertTrie(TrieNode * root,const string & word) {
    TrieNode * node = root;
    for (auto c : word){
        if (!node->children.count(c)) {
            node->children[c] = new TrieNode();
        }
        node = node->children[c];
    }
    node->word = word;
}

class Solution {
public:
    int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    bool dfs(vector<vector<char>>& board, int x, int y, TrieNode * root, set<string> & res) {
        char ch = board[x][y];        
        if (!root->children.count(ch)) {
            return false;
        }
        root = root->children[ch];
        if (root->word.size() > 0) {
            res.insert(root->word);
        }

        board[x][y] = '#';
        for (int i = 0; i < 4; ++i) {
            int nx = x + dirs[i][0];
            int ny = y + dirs[i][1];
            if (nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size()) {
                if (board[nx][ny] != '#') {
                    dfs(board, nx, ny, root,res);
                }
            }
        }
        board[x][y] = ch;

        return true;      
    }

    vector<string> findWords(vector<vector<char>> & board, vector<string> & words) {
        TrieNode * root = new TrieNode();
        set<string> res;
        vector<string> ans;

        for (auto & word: words){
            insertTrie(root,word);
        }
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[0].size(); ++j) {
                dfs(board, i, j, root, res);
            }
        }        
        for (auto & word: res) {
            ans.emplace_back(word);
        }
        return ans;        
    }
};

// 删除被匹配的单词
struct TrieNode {
    string word;
    unordered_map<char, TrieNode *> children;
    TrieNode() {
        this->word = "";
    }   
};

void insertTrie(TrieNode * root, const string & word) {
    TrieNode * node = root;
    for (auto c : word) {
        if (!node->children.count(c)) {
            node->children[c] = new TrieNode();
        }
        node = node->children[c];
    }
    node->word = word;
}

class Solution {
public:
    int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    bool dfs(vector<vector<char>>& board, int x, int y, TrieNode * root, set<string> & res) {
        char ch = board[x][y];        
        if (!root->children.count(ch)) {
            return false;
        }
        root = root->children[ch];
        if (root->word.size() > 0 ) {
            res.insert(root->word);
            root->word = "";
        }

        board[x][y] = '#';
        for (int i = 0; i < 4; ++i) {
            int nx = x + dirs[i][0];
            int ny = y + dirs[i][1];
            if (nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size()) {
                if (board[nx][ny] != '#'){
                    dfs(board, nx, ny, root,res);
                }
            }
        }
        board[x][y] = ch;
        return true;      
    }

    vector<string> findWords(vector<vector<char>> & board, vector<string> & words) {
        TrieNode * root = new TrieNode();
        set<string> res;
        vector<string> ans;

        for (auto & word: words) {
            insertTrie(root,word);
        }
        for (int i = 0; i < board.size(); ++i) {
            for(int j = 0; j < board[0].size(); ++j) {
                dfs(board, i, j, root, res);
            }
        }        
        for (auto & word: res) {
            ans.emplace_back(word);
        }
        return ans;        
    }
};
